If the different zeroes a1,a2, …,ansubscripta1subscripta2normal-…subscriptana_{1},a_{2},\,\ldots,a_{n} of Q⁢(s)QsQ(s)
have the multiplicitiesm1,m2, …,mnsubscriptm1subscriptm2normal-…subscriptmnm_{1},m_{2},\,\ldots,m_{n}, respectively,
we denote Fj⁢(s):=(s-aj)mj⁢P⁢(s)/Q⁢(s)assignsubscriptFjssuperscriptssubscriptajsubscriptmjPsQsF_{j}(s):=(s\!-\!a_{j})^{{m_{j}}}P(s)/Q(s); then

But since Q′⁢(s)=dd⁢s⁢((s-aj)⁢Qj⁢(s))=Qj⁢(s)+(s-aj)⁢Qj′⁢(s)superscriptQnormal-′sddsssubscriptajsubscriptQjssubscriptQjsssubscriptajsuperscriptsubscriptQjnormal-′sQ^{{\prime}}(s)=\frac{d}{ds}((s\!-\!a_{j})Q_{j}(s))=Q_{j}(s)\!+\!(s\!-\!a_{j})%
Q_{j}^{{\prime}}(s),
we see that Q′⁢(aj)=Qj⁢(aj)superscriptQnormal-′subscriptajsubscriptQjsubscriptajQ^{{\prime}}(a_{j})=Q_{j}(a_{j}); thus the equation (5) may
be written